A spherical ball of radius 1 times 10^{-4} ,m | vedclass

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(C) Given: Radius $r = 1 	imes 10^{-4} \,m$, Density of ball $\sigma = 10^4 \,kg \,m^{-3}$, Density of water $\delta = 10^3 \,kg \,m^{-3}$, Viscosity

Vedclass · Accepted Answer

$20.4$ m

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(C) Given: Radius $r = 1 	imes 10^{-4} \,m$, Density of ball $\sigma = 10^4 \,kg \,m^{-3}$, Density of water $\delta = 10^3 \,kg \,m^{-3}$, Viscosity of water $\eta = 9.8 	imes 10^{-4} \,Pa \cdot s$ (standard value).The terminal velocity $v$ of the ball in water is given by Stokes' Law:$v = \frac{2}{9} \frac{g r^2}{\eta} (\sigma - \delta)$Substituting the values:$v = \frac{2}{9} 	imes \frac{9.8 	imes (10^{-4})^2}{9.8 	imes 10^{-4}} 	imes (10^4 - 10^3)$$v = \frac{2}{9} 	imes 10^{-4} 	imes 9000 = 20 \,m/s$Since the velocity does not change upon entering the water, the velocity acquired after falling through height $h$ must be equal to the terminal velocity $v$.Using the equation of motion $v^2 = 2gh$:$h = \frac{v^2}{2g} = \frac{20^2}{2 	imes 9.8} = \frac{400}{19.6} \approx 20.4 \,m$.

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^4 \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If the velocity of the ball does not change after entering the water, then the value of $h$ is:

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